Mathematical Indunction (M.I) Exerxise wise Solutions Mathematical Indunction: It is a technique for proving results or establishing statements for natural numbers. This part illustrates the method through a variety of examples. Definition: Mathematical Induction is a mathematical technique which is used to prove a statement, a formula or a theorem is true for every natural …
2. Mathematical Indunction (M.I) Exerxise wise Solutions Read More »
The post 2. Mathematical Indunction (M.I) Exerxise wise Solutions appeared first on Basics In Maths.
]]>Mathematical Indunction: It is a technique for proving results or establishing statements for natural numbers. This part illustrates the method through a variety of examples.
Definition:
Mathematical Induction is a mathematical technique which is used to prove a statement, a formula or a theorem is true for every natural number.
The technique involves steps to prove a statement, as stated below −
Let P(n) or S(n) be the given statement
Step 1: For n =1
we get LHS = RHS
then P(n) is true for n= 1
Step 2: Let us assume thet P(n) is true for n =k
Step 3: We have to Prove P(n) is true for n= k + 1
Laplace:
Laplace was a mathematecian and astronomer whose work was pivotal to the development of mathematical astronomy. His most outstanding work was done in the fields of celestial mechonics, probability, differential equations, and geodesy. His five volume work on celestial mechonics earned him the title of the Newton of France.
“Analysis and natural philosophy owe their most important discoveries to this fruitful means, which is called indunction” – Pierr Simon de Laplace
Let p(n) be the given statement that
1^{2} + 2^{2} + 3^{2} + …… + n^{2}=
For n= 1
LHS = 1^{2} = 1
RHS = = = 1
LHS = RHS
P(n) is true for n = 1
Let us assume that P(n) is true for n = k
i.e., 1^{2} + 2^{2} + 3^{2} + …… + k^{2}= ………… (1)
for n = k + 1
add (k +1)^{2} on both sides of (1)
1^{2} + 2^{2} + 3^{2} + …… + k^{2} + (k +1)^{2} =
P(n) is true for n = k+ 1
∴ By the principle of M.I. P(n) is true for all n ∈ N
∴ 1^{2} + 2^{2} + 3^{2} + …… + n^{2}=
First factors of given series are: 2, 3, 4, 5, …
a = 2, d = 1
a_{n} = a + (n – 1) d
= 2 + (n – 1) (1)
= 2 + n – 1
= n + 1
Second factors of given series are: 3, 4, 5,…
a = 3, d = 1
a_{n} = 3 + (n – 1) d
= 3 + (n – 1) (1)
= 3 + n – 1
= n + 2
n^{th} term of given series is (n + 1) (n + 2)
let P(n) be the given statement that
2.3 + 3.4 + 4.5 + ……… + (n + 1) (n + 2) =
For n = 1
LHS = 2.3 = 6
RHS = = = = 6
LHS = RHS
P(n) is true for n = 1
Let us assume that P(n) is true for n = k
i.e., 2.3 + 3.4 + 4.5 + ……… + (k + 1) (k + 2) = ………… (1)
for n = k + 1
add (k + 2) (k + 3) on both sides of (1)
2.3 + 3.4 + 4.5 + ……… + (k + 1) (k + 2) + (k + 2) (k + 3)
= + (k + 2) (k + 3)
P(n) is true for n = k+ 1
∴ By the principle of M.I.
P(n) is true for all n ∈ N
∴ 2.3 + 3.4 + 4.5 + ……… up to n terms =
Sol:
let P(n) be the given statement that
For n = 1
LHS = =
RHS = = =
LHS = RHS
P (n) is true for n = 1
Let us assume that P(n) is true for n = k
………… (1)
For n = k + 1
Add on both sides of (1)
P (n) is true for n = k + 1
∴ By the principle of M.I. P(n) is true for all n ∈ N
∴
The post 2. Mathematical Indunction (M.I) Exerxise wise Solutions appeared first on Basics In Maths.
]]>Functions Exercise 1c Solutions The famous mathematician ” Lejeune Dirichlet” defined a function. Function: A variable is a symbol which represents any one of a set of numbers, if two variables x and y so related that whenever a value is assigned to x there is autometically assigned by some rule or correspondence a value to …
Functions Exercise 1c Solutions ||TS|| Read More »
The post Functions Exercise 1c Solutions ||TS|| appeared first on Basics In Maths.
]]>The famous mathematician ” Lejeune Dirichlet” defined a function.
Function: A variable is a symbol which represents any one of a set of numbers, if two variables x and y so related that whenever a value is assigned to x there is autometically assigned by some rule or correspondence a value to y, then we say y is a function of x.
Chapter 1 Functions Exercise 1c Solutions for inter first year students, prepared by Mathematics expert of www.basicsinmaths.com
Exercise 1(c) Solutions
(i)
f (x) =
given function is f (x) =
f (x) is defined when (x^{2} – 1) (x + 3) ≠ 0
⟹ (x^{2} – 1) ≠ 0 or (x + 3) ≠ 0
⟹ (x + 1) (x – 1) ≠ 0 or (x + 3) ≠ 0
⟹ x ≠ 1, x ≠ – 1 or x ≠ – 3
∴ Domain of f(x) is R – {– 1, – 3, 1}
(ii)
f (x) =
Given function is f (x) =
f (x) is defined when (x – 1) (x – 2) (x – 3) ≠ 0
⟹ x ≠ 1, x ≠ 2 or x ≠ 3
∴ Domain of f(x) is R – {1, 2, 3}
(iii)
f (x) =
Given function is f (x) =
f (x) is defined when 2 – x > 0 and 2 – x ≠ 1
⟹ 2 > x and 2 – 1 ≠ x
⟹ 2 > x and x ≠ 1
∴ Domain of f(x) is (– ∞, 2) – {1}
(iv)
f (x) =
Given function is f (x) =
f (x) is defined when x ∈ R
∴ Domain of f(x) is R
Functions Exercise 1c
(v)
f (x) =
Given function is f (x) =
f (x) is defined when 4x – x^{2} ≥ 0
⟹ x (4 – x) ≥ 0
⟹ x (x – 4) ≤ 0
⟹ (x – 0) (x – 4) ≤ 0
⟹ x ∈ [0, 4]
∴ Domain of f(x) is [0, 4]
(vi)
f (x) =
Given function is f (x) =
f (x) is defined when 1 – x^{2} > 0
⟹ x^{2} – 1 < 0
⟹ (x – 1) (x + 1) < 0
⟹ x ∈ (– 1, 1)
∴ Domain of f(x) is (– 1, 1)
(vii)
f (x) =
Given function is f (x) =
f (x) is defined when x + 1≠ 0
⟹ x ≠ – 1
∴ Domain of f(x) is R – {– 1}
(viii)
f(x) =
Given function is f (x) =
f (x) is defined when x^{2} – 25 ≥ 0
⟹ (x – 5) (x + 5) ≥ 0
⟹ x ∈ (–∞, –5] ∪ [5, ∞)
⟹ x ∈ R – (– 5, 5)
∴ Domain of f(x) is R – (– 5, 5)
Functions Exercise 1c
(ix)
f(x) =
Given function is f (x) =
f (x) is defined when x – [x] ≥ 0
⟹ x ≥ [x]
⟹ x ∈ R
∴ Domain of f(x) is R
(x)
f(x) =
Given function is f (x) =
f (x) is defined when [x] – x ≥ 0
⟹ [x] ≥ x
⟹ x ∈ Z
∴ Domain of f(x) is Z
Functions Exercise 1c
(i)
f(x) =
Given function is f (x) =
Let y =
⟹ |4 – x^{2}| = e^{y}
∵ e^{y} > 0 ∀ y ∈ R
∴ Range of f(x) is R
(ii)
f(x) =
Given function is f (x) =
f (x) is defined when [x] – x ≥ 0
⟹ [x] ≥ x
⟹ x ∈ Z
Domain of f(x) is Z
Range of f = {0}
(iii)
f(x) =
Given function is f (x) =
f (x) is defined when x ∈ R
Domain of f(x) is R
For x ∈ R [x] is an integer
Since sin nπ = 0, ∀ n ∈ z
⟹ sin π[x] = 0
∴ Range of f = {0}
(iv)
f (x) =
Given function is f (x) =
f (x) is defined when x – 2 ≠ 0
⟹ x ≠ 2
Domain of f(x) is R – {2}
Let y =
=
= x + 2
If x = 2 ⟹ y = 2 + 2 = 4
∴ Range of f(x) is R – {4}
Functions Exercise 1c
(v)
f (x) =
let y =
y^{2} = 9 + x^{2}
^{ }x^{2} = y^{2} – 9
x =
it is defined when y^{2} – 9 ≥ 0
⟹ (y – 3) (y + 3) ≥ 0
y ∈ (– ∞, – 3] ∪ [3, ∞)
but y = ≥ 0
∴ Range of f(x) is [3, ∞)
Sol:
Given f and g are real valued functions f(x) = 2x – 1 ang g (x) = x^{2}
(i)
(3f – 2g) (x) = 3 f(x) – 2g (x)
= 3 (2x – 1) – 2(x^{2})
= 6x – 3 – 2x^{2}
= – 2x^{2} + 6x – 3
∴ (3f – 2g) (x) =– 2x^{2} + 6x – 3
(ii)
(fg) (x) = f (x) g (x)
= (2x – 1) (x^{2})
= 2x^{3} + x^{2}
∴ (fg) (x) = 2x^{3} + x^{2}
(iii)
Sol:
Given f = {(1, 2), (2, – 3) (3, – 1)}
(i)
(2f) (1) = 2 f (1) = 2 × 2 = 4
(2f) (2) = 2 f (2) = 2 × – 3 = – 6
(2f) (3) = 2 f (3) = 2 × – 1 = – 2
∴ 2f = {(1, 4), (2, – 6) (3, – 2)}
(ii)
(2 + f) (1) = 2 + f (1) = 2 + 2 = 4
(2 + f) (2) = 2 + f (2) = 2 + (– 3) = – 1
(2 + f) (3) = 2 + f (3) = 2 + (– 1) = 1
∴ 2 + f = {(1, 4), (2, – 1) (3, 1)}
(iii)
(f^{2}) (1) = [f (1))]^{2} = 2^{2} = 4
(f^{2}) (2) = [f (2))]^{2} = (– 3)^{2} = 9
(f^{2}) (3) = [f (1))]^{2} = (– 1)^{2} = 1
∴ f^{2}= {(1, 4), (2, 9) (3, 1)}
(iv)
(i)
f (x) =
f(x) is defined when x^{2} – 3x + 2 ≥ 0
x^{2} – 2x – x + 2 ≥ 0
x (x – 2) – 1(x – 2) ≥ 0
(x – 1) (x – 2) ≥ 0
x ∈ (– ∞, 1] ∪ [2, ∞)
∴ Domain of f(x) is R – (1, 2)
(ii)
f(x) = log (x^{2} – 4x + 3)
f(x) is defined when x^{2} – 4x + 3 > 0
x^{2} – 3x – x + 3 > 0
x (x – 3) – 1(x – 3) > 0
(x – 1) (x – 3) > 0
x ∈ (– ∞, 1) ∪ (3, ∞)
∴ Domain of f(x) is R – [1, 3]
(iii)
f(x) =
f(x) is defined when 2 + x ≥ 0, 2 – x ≥ 0 and x ≠ 0
x ≥ – 2, x ≤ 2 and x ≠ 0
– 2 ≤ x ≤ 2 and x ≠ 0
x ∈ [–2, 2] – {0}
∴ Domain of f(x) is [–2, 2] – {0}
(iv)
f(x) =
f(x) is defined in two cases as follows:
case (i) 4 – x^{2} ≥ 0 and [x] + 2 > 0
x^{2} – 4 ≤ 0 and [x] > –2
(x – 2) (x + 2) ≤ 0 and [x] > –2
x ∈ [– 2, 2] and x ∈ [– 1, ∞)
x ∈ [– 1, 2]
case (ii) 4 – x^{2} ≤ 0 and [x] + 2 < 0
x^{2} – 4 ≥ 0 and [x] < –2
(x – 2) (x + 2) ≥ 0 and [x] < –2
x ∈ (– ∞, –2] ∪ [2, ∞) and x ∈ (–∞, –2)
x ∈ (–∞, –2)
from case (i) and case (ii)
x ∈ (–∞, –2) ∪ [– 1, 2]
∴ Domain of f(x) is (–∞, –2) ∪ [– 1, 2]
(v)
f(x) =
f(x) is defined when ≥ 0 and x – x^{2} > o
x – x^{2} ≥ (0.3)^{0} and x^{2} – x < 0
x – x^{2} ≥ 1 and x (x – 1) < 0
x^{2 }–x + 1 ≤ 0 and (x – 0) (x – 1) < 0
it is true for all x ∈ R and x ∈ (0, 1)
∴ domain of f(x) = R∩ (0, 1) = (0, 1)
(vi)
f(x) =
f(x) is defined when x +|x| ≠ 0
|x| ≠ – x
|x| ≠ – x
⟹|x| = x
⟹ x > 0
x ∈ (0, ∞)
∴ Domain of f(x) is (0, ∞)
Sol:
Given f(x) =
(i)
f (x) =
Given f(x) =
Since [x] is an integer
sin π[x] = tan π[x] = 0 ∀ x ∈ R
∴ domain of f(x) is R
and
since tan π[x] = 0
Range of f(x) = {0}
(ii)
f(x) =
Given f (x) =
It is defined when 2 – 3x ≠ 0
⟹ 2 ≠ 3x
⟹ x ≠ 2/3
∴ Domain of f (x) = R – {2/3}
Let y = f(x)
y =
⟹ y (2 – 3x) = x
2y – 3xy = x
2y = x + 3xy
2y = x (1 + 3y)
⟹ x =
It is defined when 1 + 3y ≠ 0
1 ≠ –3y
y ≠ – 1/3
∴ Range of f (x) = R – {– 1/3}
(iii)
f(x) = |x| + | 1 + x|
Given function is f (x) = |x| + |1 + x|
f (x) is defined for all x ∈ R
∴ domain of f(x) = R
f (– 3) = |– 3| + |1 – 3|
=|– 3| + |– 2|
= 3 + 2 = 5
f (– 2) = |– 2| + |1 – 2|
=|– 2| + |– 1|
= 2 + 1 = 3
f (– 1) = |– 1| + |1 – 1|
=|– 1| + |0|
= 1 + 0 = 1
f (0) = |0| + |1 + 0| = 1
f (1) = |1| + |1 + 1|
= 1 + |2|
= 1 + 2 = 3
f (2) = |2| + |1 + 2|
= |2| + |3|
= 2 + 3 = 5
f (3) = |3| + |1 + 3|
= |3| + |4|
= 3 + 4 = 7
∴ Range of f(x) = [1, ∞)
The post Functions Exercise 1c Solutions ||TS|| appeared first on Basics In Maths.
]]>Functions Exercise 1b Solutions The famous mathematician ” Lejeune Dirichlet” defined a function. Function: A variable is a symbol which represents any one of a set of numbers, if two variables x and y so related that whenever a value is assigned to x there is autometically assigned by some rule or correspondence a value to …
Functions Exercise 1b Solutions Read More »
The post Functions Exercise 1b Solutions appeared first on Basics In Maths.
]]>The famous mathematician ” Lejeune Dirichlet” defined a function.
Function: A variable is a symbol which represents any one of a set of numbers, if two variables x and y so related that whenever a value is assigned to x there is autometically assigned by some rule or correspondence a value to y, then we say y is a function of x.
Chapter 1 Functions Exercise 1b Solutions for inter first year students, prepared by Mathematics expert of www.basicsinmaths.com
Exercise 1(b) Solutions
Sol:
Given f (x) = e^{x} and g(x) =
(fog) (x) = f (g (x))
= f ( )
=
= x ————- (1)
(gof) (x) = g (f (x))
= g (e^{x})
=
= x
= x ————- (2)
From (1) and (2) f og = gof
let y = f(x)
x = f^{-1}(y)
y = e^{x}
^{ } x =
f^{-1} (x) =
let g(x) = z
x = g^{-1}(z)
z =
x = e^{z}
g^{-1 }(x) = e^{x}
Sol:
Given f (y) = , g (y) =
Now
fog(y) = f(g(y))
∴ fog(y) = y
Sol:
Given f: R ⟶ R is, g: R ⟶ R is defined by f(x) = 2x^{2} + 3 ang g (x) = 3x – 2
(i) (fog) (x) = f(g(x))
= f(3x – 2)
= 2 (3x – 2)^{2} + 3
= 2 (9x^{2} – 12x + 4) + 3
= 18 x^{2} – 24x + 8 + 3
= 18 x^{2} – 24x + 11
∴ (fog) (x) = 18 x^{2} – 24x + 11
(ii) (gof) (x) = g (f (x))
= g (2x^{2} + 3)
= 3(2x^{2} + 3) – 2
= 6x^{2} + 9 – 2
= 6 x^{2} + 7
∴ (gof) (x) = 6 x^{2} + 7
PDF Files || Inter Maths 1A &1B || (New) |
(iii) (fof) (0) = f (f (0))
= f (2(0)^{2} + 3)
= f (2(0) + 3)
= f (3)
= 2 (3)^{2} + 3
= 2 (9) + 3
= 18 + 3 = 21
(iv) go(fof) (3) = go (f (f (3)))
= go (f (2 (3)^{2} + 3))
= go (f (21))
= g (f (21))
= g (2 (21)^{2} + 3))
= g (2 (441) + 3))
= g (882 + 3)
= g (885)
= 3 (885) – 2
= 2655 – 2
= 2653
∴ go(fof) (3) = 2653
Functions Exercise 1b
Sol:
Given f: R ⟶ R; g: R ⟶ R is defined by f(x) = 3x – 1, g (x) = x^{2} + 1
(i) (fof) (x^{2} + 1) = f (f (x^{2} + 1))
= f (3 (x^{2} + 1)– 1)
= f (3x^{2} + 3– 1)
= f (3x^{2} + 2)
= 3(3x^{2} + 2) – 1
= 9x^{2} + 6 – 1
= 9x^{2} + 5
(ii) (fog) (2) = f (g (2))
= f (2^{2} + 1)
=f (4 + 1)
= f (5)
= 3(5) – 1
= 15 – 1
= 14
(iii) (gof) (2a – 3) = g (f (2a – 3))
= g (3(2a – 3) – 1)
= g (6a – 9 – 1)
= g (6a – 10)
= (6a – 10)^{2} + 1
= 36^{2} – 120a + 100 + 1
= 36^{2} – 120a + 101
Sol:
Given f(x) = and g(x) = for all x ∈ (0, ∞)
(gof) (x) = g (f (x))
= g ( )
=
∴ (gof) (x) =
Sol:
Given f(x) = 2x – 1 and g(x) = for all x ∈ R
∴ gof(x) = x
Sol:
Given f (x) = 2, g (x) = x^{2} and h(x) = 2x ∀ x ∈ R
(fo(goh)) (x) = fo (g (h (x))
= fo g(2x)
= f (g(2x))
= f((2x)^{2})
= f(4x^{2})
= 2
∴ (fo(goh)) (x) = 2
Functions Exercise 1b Solutions
(i) a, b ∈ R, f: R ⟶ R defined by f(x) = ax + b, (a ≠ 0)
Given a, b ∈ R, f: R ⟶ R defined by f(x) = ax + b
Let y = f(x)
x = f^{-1}(y)
now y = ax + b
ax = y – b
x =
f^{-1}(y) =
∴ f^{-1}(x) =
(ii) f: R ⟶ (0, ∞) defined by f(x) = 5^{x}
Let y = f(x)
x = f^{-1}(y)
now y = 5^{x}
x =
f^{-1}(y) =
∴ f^{-1}(x) =
(iii) f: (0, ∞) ⟶ R defined by f(x) =
Let y = f(x)
x = f^{-1}(y)
now y =
x = 2^{y}
f^{-1}(y) = 2^{y}
∴ f^{-1}(x) = 2^{x}
Sol:
Given, f(x) = 1 + x + x^{2} + … for
1 + x + x^{2} + … is an infinite G.P
a = 1, r = x
S_{∞} =
=
Now
f (x) =
Let y = f(x)
x = f^{-1}(y)
now y =
1 – x =
x = 1 –
=
f^{-1}(y) =
∴ f^{-1}(x) =
Sol:
Given f: [1, ∞) ⟶ [1, ∞) defined by f (x) = 2^{x (x – 1)}
Let y = f(x)
x = f^{-1}(y)
now y = 2^{x (x – 1)}
= x (x – 1)
x^{2} – x =
x^{2} – x – = 0
Functions Exercise 1b Solutions
Sol:
Given f (x) = , x ≠ ± 1
Let y = f(x)
x = f^{-1}(y)
now y =
y (x + 1) = x – 1
xy + y = x – 1
1 + y = x – xy
1 + y = x (1 – y)
x =
f^{-1}(y) =
∴ f^{-1}(x) =
Now
Sol:
Given A = {1, 2, 3}, B = {α, β, γ}, C = {p, q, r} and f: A ⟶ B; g: B ⟶ C is defined by f = {(1, α), (2, γ), (3, β)}, g = {(α, p), (β, r), (γ, p)}
A = {1, 2, 3}, B = {α, β, γ}
f: A ⟶ B; and f = {(1, α), (2, γ), (3, β)}
⟹ f (1) = α; f (2) = γ; f (3) = β
⟹ Every element of set A has a unique image in set B
f is injection (one – one)
range of f = codomain of f
⟹ f is surjection (on to)
∴ f is bijective
B = {α, β, γ}, C = {p, q, r}
g: B ⟶ C is defined by g = {(α, q), (β, r), (γ, p)}
⟹ g (α) =q; g (β) = r; g (γ) = p
⟹ Every element of set B has a unique image in set C
g is injection (one – one)
range of g = codomain of g
⟹ g is surjection (on to)
∴ g is bijective
Now
f = {(1, α), (2, γ), (3, β)}, g = {(α, q), (β, r), (γ, p)}
gof = {(1, q), (2, p), (3, r)
(gof)^{-1} = {(q, 1), (p, 2), (r, 3)} ———– (1)
f^{-1} = {(α, 1), (γ, 2), (β, 3)
g^{-1} = {(q, α), (r, β), (p, γ)}
f^{-1}og^{-1} = {(q, 1), (p, 2), (r, 3)} ———– (2)
From (1) and (2)
(gof)^{-1} = f^{-1}og^{-1}
Sol:
Given f: R ⟶ R; g: R ⟶ R is defined by f (x) = 3x – 2 and g(x) = x^{2} + 1
Let y = f(x)
x = f^{-1}(y)
now y = 3x – 2
3x = y + 2
x =
f^{-1}(y) =
f^{-1}(x) =
∴ (gof^{-1}) (2) =
(ii) (gof) (x – 1) = g (f (x – 1))
= g (3(x – 1) – 2)
= g (3x – 3 – 2)
= g (3x – 5)
= (3x – 5)^{2} + 1
= 9x^{2} – 30x + 25 + 1
= 9x^{2} – 30x + 26
∴(gof) (x – 1) = 9x^{2} – 30x + 26
Sol:
Given f = {(1, a), (2, c), (4, d), (3, b)} and g^{-1} = {(2, a), (4, b), (1, c), (3, d)}
⟹ f ^{-1} = {(a, 1), (c, 2), (d, 4), (b, 3)} and g = {(a, 2), (b, 4), (c, 1), (d, 3)}
Now gof = {(1, 2), (2, 1), (4, 3), (3, 4)}
(gof)^{-1} = {(2, 1), (1, 2), (3, 4), (4, 3)} ———(1)
g^{-1} = {(2, a), (4, b), (1, c), (3, d)} ; f ^{-1} = {(a, 1), (c, 2), (d, 4), (b, 3)}
f^{-1}og^{-1} = {(2, 1), (1, 2), (3, 4), (4, 3)} ———(2)
from (1) and (2)
(gof)^{-1} = f^{-1}og^{-1 }
Sol:
Given R ⟶ R; g: R ⟶ R is defined by f (x) = 2x – 3 and g(x) = x^{3} + 5
(fog) (x) = f (g(x))
= f (x^{3} + 5)
= 2 (x^{3} + 5) – 3
= 2 x^{3} + 10 – 3 = 2x^{3} + 7
Let y = (fog) (x)
⟹ x = (fog)^{-1}(y)
y = 2x^{3} + 7
2x^{3} = y – 7
x^{3} =
x =
(fog)^{-1}(y)=
(fog)^{-1}(x)=
Sol:
Given f(x) = x^{2}, g(x) = 2^{x}
(fog) (x) = f(g(x))
= f (2^{x})
= (2^{x})^{2}
= 2^{2x}
(gof) (x) = g(f(x))
= g (x^{2})
=
Now
(fog) (x) = (gof) (x)
⟹ 2^{2x} =
2x = x^{2} [∵ if a^{m} = a^{n} , then m = n]
x^{2} – 2x = 0
⟹ x (x – 2) = 0
⟹x = 0 or x – 2 = 0
∴ x = 0 or x = 2
Sol:
Given f (x) = , x ≠ ±1
PDF Files || Inter Maths 1A &1B || (New) 6th maths notes|| TS 6 th class Maths Concept |
The post Functions Exercise 1b Solutions appeared first on Basics In Maths.
]]>Functions Exercise 1a Solutions The famous mathematician ” Lejeune Dirichlet” defined a function. Function: A variable is a symbol which represents any one of a set of numbers, if two variables x and y so related that whenever a value is assigned to x there is autometically assigned by some rule or correspondence a value to …
Functions Exercise 1a Solutions Read More »
The post Functions Exercise 1a Solutions appeared first on Basics In Maths.
]]>The famous mathematician ” Lejeune Dirichlet” defined a function.
Function: A variable is a symbol which represents any one of a set of numbers, if two variables x and y so related that whenever a value is assigned to x there is autometically assigned by some rule or correspondence a value to y, then we say y is a function of x.
Chapter 1 Functions Exercise 1a Solutions for inter first year students, prepared by Mathematics expert of www.basicsinmaths.com
if ( function_exists( ‘pgntn_display_pagination’ ) ) pgntn_display_pagination( ‘multipage’ );
I.
then find the values of (i) f (3) (ii) f (0) (iii) f (– 1.5) (iv) f (2) + f (– 2) (v) f (– 5 )
Sol:
Given
Domain of f(x) is (– 3, ∞)
(i) f (3)
3 lies in the interval x > 1
⟹ f(x) = x + 2
f(3) = 3 + 2 = 5
∴ f (3) = 5
(ii) f (0)
0 lies in interval – 1 ≤ x ≤ 1
⟹f(x) = 2
∴ f (0) = 2
(iii) f (– 1.5)
– 1.5 lies in interval – 3 < x < – 1
⟹f(x) = x – 1
f (– 1.5) = – 1.5 – 1 = – 2.5
∴ f (– 1.5) = – 2.5
(iv) f (2) + f (– 2)
2 lies in the interval x > 1
⟹ f (x) = x + 2
f (3) = 2 + 2 = 4
f (2) = 4
– 2 lies in interval – 3 < x < – 1
⟹f(x) = x – 1
f (– 2) = – 2 – 1 = – 3
f (– 2) = – 2 – 1 = – 3
now f (2) + f (– 2) = 4 – 3 = 1
∴ f (2) + f (– 2) = 1
(v) f (– 5)
since domain of f(x) is (– 3, ∞)
f (– 5) is not defined
Sol:
Given f: R – {0} ⟶ R is defined by f(x) =
f (1/x) =
Now
f (x) + f (1/x) =
∴ f (x) + f (1/x) = 0
Sol:
Given f: R ⟶ R is defined by f(x) =
f (tan θ) =
= cos 2θ
Sol:
Given f: R – {±1} ⟶ R is defined by f(x) =
Sol:
Given A = {– 2, – 1, 0, 1, 2} and f: A ⟶ B is a surjection defined by f(x) = x^{2} + x + 1
f(– 2) = (–2)^{2} + (–2) + 1
= 4 – 2 + 1 = 3
f(– 1) = (–1)^{2} + (–1) + 1
= 1 – 1 + 1 = 1
f(0) = (0)^{2} + (0) + 1
= 0 + 0 + 1 = 1
f(1) = (1)^{2} + (1) + 1
=1 +1 + 1 = 3
f( 2) = (2)^{2} + (2) + 1
= 4 + 2 + 1 = 4
∴ B = {1, 3, 7}
Sol:
Given A = {1, 2, 3,4} and f: A ⟶ B is a surjection defined by f(x) =
Sol:
Given f (x + y) = f (xy) ∀ x, y ∈ R
Let x = 0 and y = 0
f (0 + 0) = f (0 × 0) = f (0)
f (1) = f (0 + 1)
= f (0 × 1)
= f (0)
f (2) = f (1 + 1)
= f (1 × 1)
= f (1)
= f (0)
f (3) = f (1 + 2)
= f (1 × 2)
= f (2)
= f(0)
Similarly, f(4) = 0
f(5) = 0
and so on.
∴ f is a constant function
II.
(i) f: A ⟶ A (ii) g: A ⟶ A
Sol:
A = {– 1, 0, 1}; f: A ⟶ A
f (x) = x^{2}
f (– 1) = (– 1)^{2} = 1
f (0) = (0)^{2} = 0
f (1) = (1)^{2} = 1
∵ range is not equal to co domain of f
f is nor a surjection
A = {– 1, 0, 1}; g: A ⟶ A
g (x) = x^{3}
g (– 1) = (– 1)^{3} = – 1
g (0) = (0)^{3} = 0
g (1) = (1)^{3} = 1
∵ range is equal to co domain of f
f is a surjection
let x_{1}, x_{2} ∈ R
f(x_{1}) = f(x_{2})
2x_{1} + 1 = 2x_{2} + 1
2x_{1} = 2x_{2}
x_{1} = x_{2}
x_{1}, x_{2} ∈ R, f(x_{1}) = f(x_{2}) ⟹ x_{1} = x_{2}
∴ f is an injection
Let y = f(x)
y =
3y = 2x + 1
3y – 1 = 2x
⟹ x = ∈ R
Now
∴ f is surjection
f is injection and surjection
∴ f is a bijection
let x_{1}, x_{2} ∈ R
f(x_{1}) = f(x_{2})
x_{1} = x_{2}
x_{1}, x_{2} ∈ R, f(x_{1}) = f(x_{2}) ⟹ x_{1} = x_{2}
∴ f is an injection
Let y = f(x)
y = 2^{x}
x = ∈ (0, ∞)
Now f(x) = 2^{x}
=
= y
∴ f is surjection
f is injection and surjection
∴ f is a bijection
let x_{1}, x_{2} ∈ (0, ∞)
f(x_{1}) = f(x_{2})
x_{1} = x_{2}
x_{1}, x_{2} ∈ (0, ∞), f(x_{1}) = f(x_{2}) ⟹ x_{1} = x_{2}
∴ f is an injection
Let y = f(x)
y =
x = e^{y} ∈ (0, ∞)
Now f(x) =
=
= y
∴ f is surjection
f is injection and surjection
∴ f is a bijection
let x_{1}, x_{2} ∈ [0, ∞)
f(x_{1}) = f(x_{2})
x_{1} = x_{2} [∵ x_{1}, x_{2} ∈ [0, ∞)]
x_{1}, x_{2} ∈ [0, ∞), f(x_{1}) = f(x_{2}) ⟹ x_{1} = x_{2}
∴ f is an injection
Let y = f(x)
y = x^{2}
x = ∈ [0, ∞)
Now f(x) = x^{2}
=
= y
∴ f is surjection
f is injection and surjection
∴ f is a bijection
let x_{1}, x_{2} ∈ R
f(x_{1}) = f(x_{2})
x_{1} = ± x_{2} [∵ x_{1}, x_{2} ∈ R]
x_{1}, x_{2} ∈ [0, ∞), f(x_{1}) = f(x_{2}) ⟹ x_{1} ≠ x_{2}
∴ f is not an injection
Let y = f(x)
y = x^{2}
x = ∈ R
Now f(x) = x^{2}
=
= y
∴ f is surjection
f is not an injection but surjection
∴ f is not a bijection
let x_{1}, x_{2} ∈ R
f(x_{1}) = f(x_{2})
x_{1} = ± x_{2} [∵ x_{1}, x_{2} ∈ R]
x_{1}, x_{2} ∈ [0, ∞), f(x_{1}) = f(x_{2}) ⟹ x_{1} ≠ x_{2}
∴ f is not an injection
f (1) = 1^{2} = 1
f (– 1) = (–1)^{2} = 1
here ‘– 1’ has no pre image
∴ f is not a surjection
f is not an injection and not surjection
∴ f is not a bijection
hai
Sol:
Given A = {1, 2, 3, 4}, B = {1, 3, 5, 7} and g = {(1, 1), (2, 3), (3, 5), (4, 7)}
g (1) = 1; g(2) = 3; g(3) = 5 ; g(4) = 7
here, every element of set A has a unique image in set B
∴ g: A ⟶ B is a function
And also given g(x) = ax + b
g (1) = 1
⟹ a (1) + b = 1
a + b = 1
b = 1 – a _______________ (1)
g (2) = 3
a (2) + b = 3
2a + b = 3
2a + 1 – a = 3 (from (1))
a+ 1 = 3
a = 2
b = 1 – 2
b = – 1
∴ a = 2, b = – 1
Sol:
Given the function f: R ⟶ R defined by f(x) =
= 2 f (x) f (y)
5. If the function f: R ⟶ R defined by f(x) = , then show that f (1 – x) = 1 – f (x)
and hence reduce the value of
Sol:
Given the function f: R ⟶ R defined by f(x) =
∴ f (1 – x) = 1 – f(x)
Sol:
Given the function f: {– 1, 1} ⟶ {0, 2}, defined by f(x) = ax + b is a surjection
Case(i)
If f(– 1) = 0 and f(1) = 2
a (– 1) + b = 0
– a + b = 0
b = a ————(1)
and
a (1) + b = 2
a + b = 2
a + a = 2 [ from (1)]
2a = 2
a = 1
b = 1
Case (ii)
If f(– 1) = 2 and f(1) = 0
a (– 1) + b = 2
– a + b =
b = 2 + a ————(2)
and
a (1) + b = 0
a + b = 0
a + 2 + a = 0 [ from (2)]
2a + 2 = 0
2a = – 2
a = – 1
b = 1
From Case(i) and Case (ii) a = ±1 and b = 1
Sol
Given f(x) = cos (log x)
cos (log x) cos (log y) – [cos (log x) cos (log y) – sin (log x) sin (log x)
+ cos (log x) cos (log y) + sin (log x) sin (log x)]
= cos (log x) cos (log y) – [2cos (log x) cos (log y)]
= cos (log x) cos (log y) – cos (log x) cos (log y)
∴ = 0
The post Functions Exercise 1a Solutions appeared first on Basics In Maths.
]]>TS Polycet || Solved Previous Question Papers 2020 Math The State Board of Technical Education and Training (SBTET), Telangana, Hyderabad will conduct “Polytechnic Common Entrance Test (POLYCET)” for the candidates seeking admission in to all Diploma Courses in Engineering /Non Engineering Technology. The Main Subjects in this Exam are Maths, Physics, Chemistry and Biology. Here …
Maths TS Polycet || Solved Previous Question Papers 2020 Read More »
The post Maths TS Polycet || Solved Previous Question Papers 2020 appeared first on Basics In Maths.
]]>
1.If 7 divides a^{2} then
a^{2} ను 7 భాగించినచో
(1) 7 divides a (a ను 7 భాగిస్తుంది)
(2) 7 divides ( ను 7 భాగిస్తుంది)
(3) a divide 7 (7 ను a భాగిస్తుంది)
(4) none (ఏదీ కాదు)
Answer: (1)
2. In the formula , which of the following is true?
అయిన, ఈ క్రింది వాటిలో ఏది సత్యము.
(1) x > 0, y > 0, a = 1
(2) x < 0, y < 0, a = 1
(3) a > 0, y > 0, x = 1
(4) x > 0, y > 0, a ≠ 1
Answer: (4)
Answer: (3)
Answer: (4)
Answer: (2)
TS 10th class maths concept (E/M)
TS 10th Class Maths Concept (T/M)
Page 1 of 14 |
The post Maths TS Polycet || Solved Previous Question Papers 2020 appeared first on Basics In Maths.
]]>TS Polycet || Solved Previous Question Papers 2021 Maths The State Board of Technical Education and Training (SBTET), Telangana, Hyderabad will conduct “Polytechnic Common Entrance Test (POLYCET)” for the candidates seeking admission in to all Diploma Courses in Engineering /Non Engineering Technology. The Main Subjects in this Exam are Mathematics, Physics, Chemistry and Biology. Here …
TS Polycet || Solved Previous Question Papers 2021 Maths Read More »
The post TS Polycet || Solved Previous Question Papers 2021 Maths appeared first on Basics In Maths.
]]>
Answer: (2)
Answer: (4)
(1) Natural numbers (సహజ సంఖ్య)
(2) Irrational number (కరణీయ సంఖ్య)
(3) Rational Number (అకరణీయ సంఖ్య)
(4) An Integer (పూర్ణ సంఖ్య)
Answer: (2)
Answer: (3)
Answer: (2)
<< Page 1 of 14 >>
The post TS Polycet || Solved Previous Question Papers 2021 Maths appeared first on Basics In Maths.
]]>Power point presentations About a Presentation? A communication device that relays a topic to an audience in the form of a slide show, a demonstration, a lecture or speech where words and pictures intend to complement each other. Different Types of Power Point Presentations: 1) Informative Presentations 2) Instructive Presentations 3) Persuasive Presentations 4) …
Power point Presentations Read More »
The post Power point Presentations appeared first on Basics In Maths.
]]>Power point presentations
A communication device that relays a topic to an audience in the form of a slide show, a demonstration, a lecture or speech where words and pictures intend to complement each other.
1) Informative Presentations
2) Instructive Presentations
3) Persuasive Presentations
4) Motivational Presentations
5) Decision-making Presentations
6) Progress Presentations
The post Power point Presentations appeared first on Basics In Maths.
]]>TS Tet 2022 syllabus For Mathametic Paper 2 is very useful for writing candidates for TS TET. Knowing the syllabus will make the practice of the candidates easier. TS Tet 2022 Syllabus For Mathametic Paper 2 TS TET రాసె అభ్యర్థులు చాలా ఉపయోగపడుతుంది. సిలబస్ తెలియడం వల్ల అభ్యర్థుల అభ్యసన సులువు అవుతుంది. free Syllabus Telangana Teachers …
TS TET syllabus 2022 For Mathametic Paper 2 Read More »
The post TS TET syllabus 2022 For Mathametic Paper 2 appeared first on Basics In Maths.
]]>
Telangana Teachers Elgible Test Official Website : Click Here |
1 . ప్రధాన సంఖ్యలు మరియు సంయుక్త సంఖ్యలు (Prime and Composite numbers)
2. భాజనీయత సూత్రాలు (Test of divisibility Rules)
3. సంఖ్యల రకాలు (types of numbers)
4. వాస్తవ సంఖ్యలు (Real numbers)
5. భిన్నాలు మరియు దశాంశ భిన్నాలు (Fractions and Decimal Fractions)
6. క .సా .గు . మరియు గ . సా .భా. యూక్లిడ్ భాగహార పద్ధతి (LCM and HCF – Euclid division Lemma)
7. వర్గాలు – వర్గ మూలాలు (Squares – Square roots)
8. ఘనాలు – ఘన మూలాలు (Cubs – Cube roots
9. సంఖ్యల అమరిక మరియు సంఖ్యల ఫజిల్ (Numbers Pattern and Numbers puzzle)
10. యూక్లిడ్ భాజనీయత సూత్రం (Euclid Division lemma)
11. సంవర్గమానము (Logarithms)
The post TS TET syllabus 2022 For Mathametic Paper 2 appeared first on Basics In Maths.
]]>TS Intermediate First Year Maths Question Papers 2022 PDF As per reduced syllabus were designed by the ‘Basics in Maths‘ team. These Model papers 2022 to do help the intermediate First-year Maths students. TS Inter Maths 1A and 1B Modelpapers 2022 as per reduced syllabus are very useful in IPE examinations. Ts Inter …
TS Intermediate First Year Maths Question Papers 2022 PDF Read More »
The post TS Intermediate First Year Maths Question Papers 2022 PDF appeared first on Basics In Maths.
]]>As per reduced syllabus were designed by the ‘Basics in Maths‘ team.
These Model papers 2022 to do help the intermediate First-year Maths students.
TS Inter Maths 1A and 1B Modelpapers 2022 as per reduced syllabus are very useful in IPE examinations.
TS Inter Maths 1A Model papers 2022 as per reduced syllabus were designed by the ‘Basics in Maths‘ team.
These Model papers 2022 to do help the intermediate First-year Maths students.
The post TS Intermediate First Year Maths Question Papers 2022 PDF appeared first on Basics In Maths.
]]>10 Maths ICSE Banking MCQS With Answers TS 10th class maths concept (E/M) 1. Mr Gupta gets ₹ 6,455 at the end of the one year at the rate of 14% per annum in a recurring deposit account find the monthly instalment A) ₹ 200 B) ₹ 600 C) ₹ 500 D) ₹ …
10 th Maths ICSE Banking MCQS With Answers Read More »
The post 10 th Maths ICSE Banking MCQS With Answers appeared first on Basics In Maths.
]]>
A) ₹ 200
B) ₹ 600
C) ₹ 500
D) ₹ 300
A) ₹ 7865
B) ₹ 12875
C) ₹875
D) ₹13865
A) ₹ 78650
B) ₹ 2350
C) ₹12350
D) ₹13050
The post 10 th Maths ICSE Banking MCQS With Answers appeared first on Basics In Maths.
]]>